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David Terr
Ph.D. Math, UC Berkeley

 

Home >> Pre-Calculus >> 6. Additional Topics in Trigonometry

>> 6.2. The Law of Cosines

 

6.1. The Law of Sines

There are at least two more important trigonometric theorems, namely the Law of Sines and the Law of Cosines. Both of these laws are useful in determining the angles of arbitrary triangles from the lengths of the sides. In this section we investigate the Law of Sines.

Theorem 6.1.1: Consider the triangle shown below with angles A, B, and C and opposite sides of length a, b, and c respecitvely. Then we have

  • (6.1.2) (sin A) / a = (sin B) / b = (sin C) / c

law of sines

Proof: Consider the diagram below, which is the same diagram but with the altitude included. It is clear from the geometry of the figure that the altitude is given by the formulas a sin C = c sin A. Dividing both sides by ac, we see that (sin A) / a = (sin C) / c. To get the remaining equality, we repeat this argument with one of the other two altitudes, i.e. the line segment perpendicular to the side with length a and passing through the angle of measure A or the line segment perpendicular to the side with length c and passing through the angle of measure C.

QED

law of sines proof

The law of sines can be used to determine the lengths of two sides of a triangle, given the angles and length of the remaining side with opposite angle specified.

 

Example 1: A triangle has angles with measures of 40, 60, and 80 degrees and a side of length 10 opposite the 60-degree angle. Determine the lengths of the other two sides.

Solution: We let A = 40°, B = 60°, C = 80°, and b = 10. From the Law of Sines, we have (sin A) / a = (sin B) / b, whence

a = b sin A / sin B = 10 sin 40°/ sin 60° = (10)(0.643) / 0.866 = 7.42

and

c = b sin C / sin B = 10 sin 80°/ sin 60° = (10)(0.985) / 0.866 = 11.37

 

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